Project Unsupervised Learning

Andres Delgadillo

1 Project: AllLife Bank Customer Segmentation

1.1 Objective

To identify different segments in the existing customer, based on their spending patterns as well as past interaction with the bank, using clustering algorithms, and provide recommendations to the bank on how to better market to and service these customers.

1.2 Data Dictionary

• Sl_No: Primary key of the records
• Customer Key: Customer identification number
• Average Credit Limit: Average credit limit of each customer for all credit cards
• Total credit cards: Total number of credit cards possessed by the customer
• Total visits bank: Total number of visits that customer made (yearly) personally to the bank
• Total visits online: Total number of visits or online logins made by the customer (yearly)
• Total calls made: Total number of calls made by the customer to the bank or its customer service department (yearly)

1.3 Questions to be answered

• In how many groups the bank can classify the customers?
• What are the main characteristics of different groups of customers?

2 Import packages and turnoff warnings

import warnings
warnings.filterwarnings("ignore")

import numpy as np
import pandas as pd
import seaborn as sns
import matplotlib.pyplot as plt
import pandas_profiling
import time
sns.set(color_codes=True)
%matplotlib inline

# to scale the data using z-score
from sklearn.preprocessing import StandardScaler

# to compute distances
from scipy.spatial.distance import cdist

# to perform k-means clustering and compute silhouette scores
from sklearn.cluster import KMeans
from sklearn.metrics import silhouette_score

# to visualize the elbow curve and silhouette scores
from yellowbrick.cluster import KElbowVisualizer, SilhouetteVisualizer

# to compute distances
from scipy.spatial.distance import pdist

# to perform hierarchical clustering, compute cophenetic correlation, and create dendrograms
from sklearn.cluster import AgglomerativeClustering
from scipy.cluster.hierarchy import dendrogram, linkage, cophenet

3 Import dataset and quality of data

# read data from excel file
data = pd.read_excel('Credit Card Customer Data.xlsx')

# get columns
data.columns

Index(['Sl_No', 'Customer Key', 'Avg_Credit_Limit', 'Total_Credit_Cards',
       'Total_visits_bank', 'Total_visits_online', 'Total_calls_made'],
      dtype='object')

# get size of dataset
data.shape

(660, 7)

# check dataset information 
data.info()

<class 'pandas.core.frame.DataFrame'>
RangeIndex: 660 entries, 0 to 659
Data columns (total 7 columns):
 #   Column               Non-Null Count  Dtype
---  ------               --------------  -----
 0   Sl_No                660 non-null    int64
 1   Customer Key         660 non-null    int64
 2   Avg_Credit_Limit     660 non-null    int64
 3   Total_Credit_Cards   660 non-null    int64
 4   Total_visits_bank    660 non-null    int64
 5   Total_visits_online  660 non-null    int64
 6   Total_calls_made     660 non-null    int64
dtypes: int64(7)
memory usage: 36.2 KB

# check dataset missing values
total = data.isnull().sum().sort_values(ascending=False) # total number of null values
print(total)

Sl_No                  0
Customer Key           0
Avg_Credit_Limit       0
Total_Credit_Cards     0
Total_visits_bank      0
Total_visits_online    0
Total_calls_made       0
dtype: int64

# check for duplicates
data.duplicated().sum()

0

This first assessment of the dataset shows:

• There are 660 rows and 7 features
• There is no missing data
• There are not duplicated rows

4 Exploratory Data Analysis

# check first rows of data
data.head()

	Sl_No	Customer Key	Avg_Credit_Limit	Total_Credit_Cards	Total_visits_bank	Total_visits_online	Total_calls_made
0	1	87073	100000	2	1	1	0
1	2	38414	50000	3	0	10	9
2	3	17341	50000	7	1	3	4
3	4	40496	30000	5	1	1	4
4	5	47437	100000	6	0	12	3

# check last rows of data
data.tail()

	Sl_No	Customer Key	Avg_Credit_Limit	Total_Credit_Cards	Total_visits_bank	Total_visits_online	Total_calls_made
655	656	51108	99000	10	1	10	0
656	657	60732	84000	10	1	13	2
657	658	53834	145000	8	1	9	1
658	659	80655	172000	10	1	15	0
659	660	80150	167000	9	0	12	2

4.1 Pandas profiling report

We can get a first statistical and descriptive analysis using pandas_profiling

# get pandas profiling report
pandas_profiling.ProfileReport(data)

Pandas Profiling report is showing some warnings/characteristics in the data:

• Sl_No is uniformly distributed, and all rows are unique values has unique values
• Customer Key has 665 distinct values. That means there are some customers with multiple data
• Total_visits_bank, Total_visits_online, and Total_calls_made have values with zeros

4.2 Univariate Analysis

# get stats for the columns
data.describe().T

	count	mean	std	min	25%	50%	75%	max
Sl_No	660.0	330.500000	190.669872	1.0	165.75	330.5	495.25	660.0
Customer Key	660.0	55141.443939	25627.772200	11265.0	33825.25	53874.5	77202.50	99843.0
Avg_Credit_Limit	660.0	34574.242424	37625.487804	3000.0	10000.00	18000.0	48000.00	200000.0
Total_Credit_Cards	660.0	4.706061	2.167835	1.0	3.00	5.0	6.00	10.0
Total_visits_bank	660.0	2.403030	1.631813	0.0	1.00	2.0	4.00	5.0
Total_visits_online	660.0	2.606061	2.935724	0.0	1.00	2.0	4.00	15.0
Total_calls_made	660.0	3.583333	2.865317	0.0	1.00	3.0	5.00	10.0

• Avg_Credit_Limit
  • Average credit limit is between 3000 and 200000
  • Mean is 34574.2 and median 18000
  • Distribution is skewed
• Total_Credit_Cards
  • Total number of credit cards is between 1 and 10
  • Mean is 4.7 and median is 5
• Total_visits_bank
  • Total number of visits is between 0 and 5
  • Mean is 2.4 and median is 2
• Total_visits_online
  • Total number of visits online is between 0 and 15
  • Mean is 2.6 and median is 2
• Total_calls_made
  • Total number of calls made is between 0 and 10
  • Mean is 3.6 and median is 3

4.3 Pairplot.

We are going to perform bivariate analysis to understand the relationship between the columns

# Continuous columns
con_col = ['Avg_Credit_Limit', 'Total_Credit_Cards','Total_visits_bank', 
           'Total_visits_online', 'Total_calls_made']
# Pairplot for continuous columns
sns.pairplot(data[con_col], diag_kind='kde');

4.4 Bivariate and Multivariate Analysis

# Get correlation matrix for numeric variables
data[con_col].corr()

	Avg_Credit_Limit	Total_Credit_Cards	Total_visits_bank	Total_visits_online	Total_calls_made
Avg_Credit_Limit	1.000000	0.608860	-0.100312	0.551385	-0.414352
Total_Credit_Cards	0.608860	1.000000	0.315796	0.167758	-0.651251
Total_visits_bank	-0.100312	0.315796	1.000000	-0.551861	-0.506016
Total_visits_online	0.551385	0.167758	-0.551861	1.000000	0.127299
Total_calls_made	-0.414352	-0.651251	-0.506016	0.127299	1.000000

# Display correlation matrix in a heatmap
fig, ax = plt.subplots(figsize=(7,5)) 
sns.heatmap(data[con_col].corr(), annot=True, ax=ax);

Features with positive correlation:

• Avg_Credit_Limit and Total_Credit_Cards have 0.61 correlation
• Avg_Credit_Limit and Total_visits_online have 0.55 correlation

Features with negative correlation

• Avg_Credit_Limit and Total_calls_made have -0.41 correlation
• Total_Credit_Cards and Total_calls_made have -0.65 correlation
• Total_visits_bank and Total_visits_online have -0.51 correlation
• Total_visits_bank and Total_calls_made have -0.51 correlation

5 Data Pre-Processing

5.1 Feature Engineering

• Drop Sl_No column
• Investigate duplicate rows for Customer Key

# Drop CustomerID column
data.drop(['Sl_No'], axis=1, inplace=True)

# Duplicated rows in Customer Key
data[data['Customer Key'].duplicated(keep=False)].sort_values(by=['Customer Key'])

	Customer Key	Avg_Credit_Limit	Total_Credit_Cards	Total_visits_bank	Total_visits_online	Total_calls_made
48	37252	6000	4	0	2	8
432	37252	59000	6	2	1	2
4	47437	100000	6	0	12	3
332	47437	17000	7	3	1	0
411	50706	44000	4	5	0	2
541	50706	60000	7	5	2	2
391	96929	13000	4	5	0	0
398	96929	67000	6	2	2	2
104	97935	17000	2	1	2	10
632	97935	187000	7	1	7	0

• There are 5 customers (37252, 47437, 50706, 96929, 97935) with duplicated information. The duplicate values might correspond to customer profile changes, and as such, there is no need to delete these records as these are actual occurrences at some point in the time
• We are going to keep all records since we do not have more information about the data gathering process

# Drop Customer Key column
data.drop(['Customer Key'], axis=1, inplace=True)

5.2 Outliers detection

We are going to analyze if there are outliers in the features

# This function create a plot with the boxplot and distribution of the series
def histboxplot(feature):
    # creating the 2 subplots
    f2, (ax_box1, ax_hist1) = plt.subplots(nrows = 2, # Number of rows of the subplot grid= 2
                                           ncols = 1, # Number of Columns of the subplot grid= 1
                                           sharex = 'col', # x-axis will be shared among columns
                                           figsize = (7,6),
                                           gridspec_kw = {"height_ratios": (.25, .75)});
    sns.boxplot(feature, ax=ax_box1, showmeans=True, color='violet'); # boxplot 
    sns.distplot(feature, kde=True, ax=ax_hist1); # histogram
    ax_hist1.axvline(np.mean(feature), color='green', linestyle='--'); # Add mean to the histogram
    ax_hist1.axvline(np.median(feature), color='black', linestyle='-');

Avg_Credit_Limit

# display histbox 
histboxplot(data['Avg_Credit_Limit'])

There are several points with average credit limit above the upper whisker 100000. However, those values could correspond to premium customers. Therefore, we are not going to clip those values

Total_Credit_Cards

# display histbox 
histboxplot(data['Total_Credit_Cards'])

There are not outliers

Total_visits_bank

# display histbox 
histboxplot(data['Total_visits_bank'])

There are not outliers

Total_visits_online

# display histbox 
histboxplot(data['Total_visits_online'])

There are some points above the upper whisker. However, they could correspond to consumers that login more times online. Therefore, we are not going to clip those values

Total_calls_made

# display histbox 
histboxplot(data['Total_calls_made'])

There are not outliers

5.3 Scaling data

# Scaling the data set before clustering
scaler = StandardScaler()
data_scaled = scaler.fit_transform(data)
# Creating a dataframe from the scaled data
data_scaled_df = pd.DataFrame(data_scaled, columns=data.columns)
data_scaled_df

	Avg_Credit_Limit	Total_Credit_Cards	Total_visits_bank	Total_visits_online	Total_calls_made
0	1.740187	-1.249225	-0.860451	-0.547490	-1.251537
1	0.410293	-0.787585	-1.473731	2.520519	1.891859
2	0.410293	1.058973	-0.860451	0.134290	0.145528
3	-0.121665	0.135694	-0.860451	-0.547490	0.145528
4	1.740187	0.597334	-1.473731	3.202298	-0.203739
...	...	...	...	...	...
655	1.713589	2.443892	-0.860451	2.520519	-1.251537
656	1.314621	2.443892	-0.860451	3.543188	-0.553005
657	2.937092	1.520613	-0.860451	2.179629	-0.902271
658	3.655235	2.443892	-0.860451	4.224968	-1.251537
659	3.522245	1.982253	-1.473731	3.202298	-0.553005

660 rows × 5 columns

6 K-means Clustering

6.1 Elbow curve

We are going to use the Elbow Curve method to identify the optimal number of clusters

# start time
start_time = time.time()

# set of different number of clusters (K)
clusters = range(1, 9)
# list with distortions
meanDistortions = []

# loop in clusters
for k in clusters:
    # create and fit KMeans model for k
    model = KMeans(n_clusters=k)
    model.fit(data_scaled_df)
    # predict cluster for each row
    prediction = model.predict(data_scaled_df)
    # calculate average distortion using euclidean distance
    distortion = (
        sum(
            np.min(cdist(data_scaled_df, model.cluster_centers_, "euclidean"), axis=1)
        )
        / data_scaled_df.shape[0]
    )
    # append average distortion
    meanDistortions.append(distortion)
    print("Number of Clusters:", k, "\tAverage Distortion:", distortion)

# plot elbow curve    
plt.plot(clusters, meanDistortions, "bx-")
plt.xlabel("k")
plt.ylabel("Average Distortion")
plt.title("Selecting k with the Elbow Method", fontsize=20)

print("\n\n--- Execution time %s seconds ---" % (time.time() - start_time))

Number of Clusters: 1 	Average Distortion: 2.0069222262503614
Number of Clusters: 2 	Average Distortion: 1.4571553548514269
Number of Clusters: 3 	Average Distortion: 1.1466276549150365
Number of Clusters: 4 	Average Distortion: 1.0463825294774465
Number of Clusters: 5 	Average Distortion: 0.9908683849620168
Number of Clusters: 6 	Average Distortion: 0.9426543606899347
Number of Clusters: 7 	Average Distortion: 0.9095950026993534
Number of Clusters: 8 	Average Distortion: 0.8902998903180782


--- Execution time 0.4954409599304199 seconds ---

• The appropriate value of k from the elbow curve seems to be 3

6.2 Silhouette Score

Now, we are going to use the Silhouette Score to identify the optimal number of clusters

# start time
start_time = time.time()

# set of different number of clusters (K)
clusters = range(2, 10)
# list with silhouette scores
sil_score = []

# loop in clusters
for k in clusters:
    # create and fit KMeans model for k
    model = KMeans(n_clusters=k)
    preds = model.fit_predict((data_scaled_df))
    # calculate silhouette_score
    score = silhouette_score(data_scaled_df, preds)
    sil_score.append(score)
    print("For n_clusters = {}, silhouette score is {}".format(k, score))

# plot silhouette_score   
plt.plot(clusters, sil_score, "bx-")
plt.xlabel("k")
plt.ylabel("silhouette_score")
plt.title("Selecting k with Silhouette Score", fontsize=20)


print("\n\n--- Execution time %s seconds ---" % (time.time() - start_time))

For n_clusters = 2, silhouette score is 0.41842496663215445
For n_clusters = 3, silhouette score is 0.5157182558881063
For n_clusters = 4, silhouette score is 0.3556670619372605
For n_clusters = 5, silhouette score is 0.2726898791817692
For n_clusters = 6, silhouette score is 0.2553892519918698
For n_clusters = 7, silhouette score is 0.24865851948404674
For n_clusters = 8, silhouette score is 0.22708508639891067
For n_clusters = 9, silhouette score is 0.21144805721637253


--- Execution time 0.3671855926513672 seconds ---

• From the silhouette scores, it seems that 3 is a good value of k

# start time
start_time = time.time()

# Clusters with silhouette coefficients for K=3
visualizer = SilhouetteVisualizer(KMeans(3, random_state=1))
visualizer.fit(data_scaled_df)
visualizer.show();

print("\n\n--- Execution time %s seconds ---" % (time.time() - start_time))

--- Execution time 0.2004995346069336 seconds ---

• From the Silhouette Plot we can see that the silhouette coefficients values for the 3 groups are high. Therefore, 3 clusters is an optimal value

6.3 Customer profile K=3

K=3 is an appropriate number for clusters because it has the highest silhouette score and there is knick at 3 in the elbow curve

# start time
start_time = time.time()

# create and fit Kmeans model
kmeans = KMeans(n_clusters=3, random_state=0)
kmeans.fit(data_scaled_df)

print("\n\n--- Execution time %s seconds ---" % (time.time() - start_time))

--- Execution time 0.01558995246887207 seconds ---

Average values for each feature

# adding kmeans cluster labels to the original dataframe
data['K_means_cluster'] = kmeans.labels_
# create cluster profile
cluster_profile_kmeans = data.groupby('K_means_cluster').mean()
# number of rows in each cluster
cluster_profile_kmeans['Count'] = (data.groupby('K_means_cluster')['Avg_Credit_Limit'].count().values)

# let's display cluster profiles
cluster_profile_kmeans.style.highlight_max(color="lightgreen", axis=0)

	Avg_Credit_Limit	Total_Credit_Cards	Total_visits_bank	Total_visits_online	Total_calls_made	Count
K_means_cluster
0	33782.383420	5.515544	3.489637	0.981865	2.000000	386
1	12174.107143	2.410714	0.933036	3.553571	6.870536	224
2	141040.000000	8.740000	0.600000	10.900000	1.080000	50

Box plots for each feature

# Create subplots for the different boxplot charts
fig, axes = plt.subplots(1, 5, figsize=(16, 6))
for col in range(5):
    # select the specific feature in con_col
    feature = con_col[col]
    # dsiplay the boxplot
    sns.boxplot(data['K_means_cluster'], data[feature], ax=axes[col], )
fig.tight_layout(pad=2.0)

7 Hierarchical Clustering

Now, we are going to use the Hierarchical Clustering method to group the customers

7.1 Cophenetic correlation

We are going to use the Cophenetic correlation to identify the appropriate distance metric and linkage method

# start time
start_time = time.time()

# list of distance metrics
distance_metrics = ['euclidean', 'chebyshev', 'mahalanobis', 'cityblock']

# list of linkage methods
linkage_methods = ['single', 'complete', 'average', 'centroid', 'ward', 'weighted']

high_cophenet_corr = 0
high_dm_lm = [0, 0]

# dataframe with Cophenetic correlation
columns = ['Metric', 'Linkage Method', 'Cophenetic correlation']#, 'Cophenetic distance']
cophenet_corr = pd.DataFrame(columns=columns)

# for each metric
for dm in distance_metrics:
    # for each linkage methos
    for lm in linkage_methods:
        # centroid and ward methods require the distance metric to be Euclidean
        if (lm=='centroid' or lm=='ward') and dm!='euclidean' :
            pass
        else:
            # calculate Cophenetic correlation
            Z = linkage(data_scaled_df, metric=dm, method=lm)
            c, coph_dists = cophenet(Z, pdist(data_scaled_df))
            cophenet_corr = cophenet_corr.append(pd.DataFrame([[dm, lm, c]], columns=columns),
                                               ignore_index = True)
print("\n\n--- Execution time %s seconds ---" % (time.time() - start_time))

# sort results by Cophenetic correlation
cophenet_corr.sort_values(by='Cophenetic correlation',ascending=False)

--- Execution time 0.300518274307251 seconds ---

	Metric	Linkage Method	Cophenetic correlation
2	euclidean	average	0.897708
8	chebyshev	average	0.897416
16	cityblock	average	0.896329
3	euclidean	centroid	0.893939
9	chebyshev	weighted	0.891362
5	euclidean	weighted	0.886175
17	cityblock	weighted	0.882552
15	cityblock	complete	0.873148
1	euclidean	complete	0.859973
7	chebyshev	complete	0.853347
12	mahalanobis	average	0.832699
13	mahalanobis	weighted	0.780599
4	euclidean	ward	0.741516
0	euclidean	single	0.739122
6	chebyshev	single	0.738235
14	cityblock	single	0.725238
10	mahalanobis	single	0.705806
11	mahalanobis	complete	0.666353

• The Cophenetic correlation is maximum with Euclidean distance and average linkage

7.2 Dendograms

Now, we are going to use the Dendograms to identify the proper number of clusters.

# start time
start_time = time.time()

# list of linkage methods
linkage_methods = ["single", "complete", "average", "centroid", "ward", "weighted"]

# lists to save results of cophenetic correlation calculation
compare_cols = ["Linkage", "Cophenetic Coefficient"]

# to create a subplot image
fig, axs = plt.subplots(len(linkage_methods), 1, figsize=(15, 30))

# We will enumerate through the list of linkage methods above
# For each linkage method, we will plot the dendrogram and calculate the cophenetic correlation
for i, method in enumerate(linkage_methods):
    Z = linkage(data_scaled_df, metric="euclidean", method=method)

    dendrogram(Z, ax=axs[i])
    axs[i].set_title(f"Dendrogram ({method.capitalize()} Linkage)")

    coph_corr, coph_dist = cophenet(Z, pdist(data_scaled_df))
    axs[i].annotate(
        f"Cophenetic\nCorrelation\n{coph_corr:0.2f}",
        (0.80, 0.80),
        xycoords="axes fraction",
    )

print("\n\n--- Execution time %s seconds ---" % (time.time() - start_time))

--- Execution time 133.59463214874268 seconds ---

• Average, centroid and weighted linkage have cophenetic correlation higher than 0.89
• For those 3 metrics the dendograms are similar
• It seems that 3 clusters are the optimal number of groups for those 3 linkage metrics.

7.3 Customer profile

• We are going to use Euclidean distance and average linkage because this combination presents the highest cophenetic correlation.
• 3 is an appropriate number for clusters from the dendogram for euclidean distance and average linkage

# start time
start_time = time.time()

HCmodel = AgglomerativeClustering(n_clusters=3, affinity="euclidean", linkage="average")
HCmodel.fit(data_scaled_df)

print("\n\n--- Execution time %s seconds ---" % (time.time() - start_time))

--- Execution time 0.009972810745239258 seconds ---

Average values for each feature

# adding kmeans cluster labels to the original dataframe
data['Hierarchical_cluster'] = HCmodel.labels_
# create cluster profile
cluster_profile_hie = data.groupby('Hierarchical_cluster').mean()
# number of rows in each cluster
cluster_profile_hie['Count'] = (data.groupby('Hierarchical_cluster')['Avg_Credit_Limit'].count().values)

# let's display cluster profiles
cluster_profile_hie.style.highlight_max(color="lightgreen", axis=0)

	Avg_Credit_Limit	Total_Credit_Cards	Total_visits_bank	Total_visits_online	Total_calls_made	K_means_cluster	Count
Hierarchical_cluster
0	33713.178295	5.511628	3.485788	0.984496	2.005168	0.002584	387
1	141040.000000	8.740000	0.600000	10.900000	1.080000	2.000000	50
2	12197.309417	2.403587	0.928251	3.560538	6.883408	1.000000	223

Box plots for each feature

# Create subplots for the different boxplot charts
fig, axes = plt.subplots(1, 5, figsize=(16, 6))
for col in range(5):
    # select the specific feature in con_col
    feature = con_col[col]
    # dsiplay the boxplot
    sns.boxplot(data['Hierarchical_cluster'], data[feature], ax=axes[col], )
fig.tight_layout(pad=2.0)

8 K-means and Hierarchical Clustering comparison

Now, we are going to compare both techniques and the clusters found in both of them

8.1 Execution time comparison and number of clusters

• Which clustering technique took less time for execution?

  • Fit the data is very fast in those techniques. Although, Hierarchical clustering is slightly faster than K-means ( Execution time 0.0065 seconds for Hierarchical clustering vs Execution time 0.0156 seconds for K-means)
  • However, the time to identify the optimal number of clusters is much higher in Hierarchical clustering, the highest execution time was to plot the Dendograms (133 seconds) while the Elbow curve and Silhouette scores took less than 1 second

• Which clustering technique gave you more distinct clusters, or are they the same?

  • 3 clusters was the optimal number for both techniques

8.2 Cluster comparison

Now, we are going to compare the clusters found in each technique

# Kmeans clustering
cluster_profile_kmeans.style.highlight_max(color="lightgreen", axis=0)

	Avg_Credit_Limit	Total_Credit_Cards	Total_visits_bank	Total_visits_online	Total_calls_made	Count
K_means_cluster
0	33782.383420	5.515544	3.489637	0.981865	2.000000	386
1	12174.107143	2.410714	0.933036	3.553571	6.870536	224
2	141040.000000	8.740000	0.600000	10.900000	1.080000	50

# Hierarchical clustering
cluster_profile_hie.style.highlight_max(color="lightgreen", axis=0)

	Avg_Credit_Limit	Total_Credit_Cards	Total_visits_bank	Total_visits_online	Total_calls_made	K_means_cluster	Count
Hierarchical_cluster
0	33713.178295	5.511628	3.485788	0.984496	2.005168	0.002584	387
1	141040.000000	8.740000	0.600000	10.900000	1.080000	2.000000	50
2	12197.309417	2.403587	0.928251	3.560538	6.883408	1.000000	223

Both techniques determine almost the same clusters.

kmeans_clusters = data['K_means_cluster'].unique()
hie_clusters = data['Hierarchical_cluster'].unique()
for kmeans_cluster in kmeans_clusters:
    for hierachical_cluster in hie_clusters:
        ix = (data['K_means_cluster']==kmeans_cluster) & (data['Hierarchical_cluster']==hierachical_cluster)
        if sum(ix)>0:
            print(f'K_means cluster: {kmeans_cluster}, Hierarchical cluster: {hierachical_cluster}, Count: {sum(ix)}')
        

K_means cluster: 0, Hierarchical cluster: 0, Count: 386
K_means cluster: 1, Hierarchical cluster: 0, Count: 1
K_means cluster: 1, Hierarchical cluster: 2, Count: 223
K_means cluster: 2, Hierarchical cluster: 1, Count: 50

• Kmeans cluster 0 corresponds to Hierarchical cluster 0
• Kmeans cluster 1 corresponds to Hierarchical cluster 2
• Kmeans cluster 2 corresponds to Hierarchical cluster 1
• The only difference is 1 customer assigned in Kmeans cluster 1 and Hierarchical cluster 0. Let's investigate the characteristics of this customer

# customer in kmeans cluster 1 and hierarchical cluster 0
ix = (data['K_means_cluster']==1) & (data['Hierarchical_cluster']==0)
data[ix]

	Avg_Credit_Limit	Total_Credit_Cards	Total_visits_bank	Total_visits_online	Total_calls_made	K_means_cluster	Hierarchical_cluster
313	7000	4	2	2	4	1	0

This customer seems more similar to Hierarchical cluster 2 than cluster 0. Therefore, Kmeans clustering performed better clustering for this specific customer

8.3 Insights about different customers

Now, we are going to plot the different characteristics of the customers

# Create subplots for the different boxplot charts
fig, axes = plt.subplots(1, 5, figsize=(16, 6))
for col in range(5):
    # select the specific feature in con_col
    feature = con_col[col]
    # dsiplay the boxplot
    sns.boxplot(data['K_means_cluster'], data[feature], ax=axes[col], )
fig.tight_layout(pad=2.0)

There are 3 different groups of customers with the following characteristics:

• Group Low Average Credit Limit and Number of Credit Cards - Medium Interaction with the bank:
  • Low Average Credit Limit (Average: 12174.1)
  • Low Total of Credit Cards (Average: 2.4)
  • Medium number of visits to the bank (Average: 0.9)
  • Medium number of visits online (Average: 3.5)
  • High number of calls made (Average: 6.8)
• Group Medium Average Credit Limit and Number of Credit Cards - High in-person / Low online Interaction with the bank :
  • Medium Average Credit Limit (Average: 33782.4)
  • Medium Total of Credit Cards (Average: 5.5)
  • High number of visits to the bank (Average: 3.5)
  • Low number of visits online (Average: 1)
  • Medium number of calls made (Average: 2)
• Group High Average Credit Limit and Number of Credit Cards - Low in-person / High online Interaction with the bank:
  • High Average Credit Limit (Average: 141040.0)
  • High Total of Credit Cards (Average: 8.7)
  • Low number of visits to the bank (Average: 0.6)
  • High number of visits online (Average: 6.9)
  • Low number of calls made (Average: 1.1)

9 Actionable Insights & Recommendations

• Both K-means and Hierarchical clustering identified 3 different groups of customers. These segments of customers are:
  • Low Average Credit Limit and Number of Credit Cards - Medium Interaction with the bank
  • Medium Average Credit Limit and Number of Credit Cards - High in-person / Low online Interaction with the bank
  • High Average Credit Limit and Number of Credit Cards - Low in-person / High online Interaction with the bank
• With these 3 different segments of customers, the bank could create personalized campaigns to target these customers and interactions:
  • Although the first segment have low average credit limit and number of credit cards. They like to interact with the bank in in-person, online and calling the bank
  • The second group have medium average credit limit and number of credit cards. This group of customers prefer to have in-person interaction with the bank
  • The third group have high average credit limit and number of credit cards. They rely mainly in online interaction.
• The bank should improve the online interaction and customer service becausecthat would improve the perception of the customers with high credit limit
• The bank should create campaigns for users with low and medium credit limit to use more online services. In-person interactions are more expensive to on-line services, and the bank could reduce cost of services for users with low and medium credit limit